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7.3: Conservation of Momentum in Two Dimensions

Difficulty Level: At Grade Created by: CK-12

In a game of billiards, it is important to be able to visualize collisions in two dimensions – the best players not only know where the target ball is going but also where the cue ball will end up.

Conservation of Momentum in Two Dimensions

Conservation of momentum in all closed systems is valid, regardless of the directions of the objects before and after they collide. Most objects are not confined to a single line, like trains on a rail. Rather, many objects, like billiard balls or cars, can move in two dimensions. Conservation of momentum for these objects can also be calculated; momentum is a vector and collisions of objects in two dimensions can be represented by axial vector components. To review axial components, revisit Vectors: Resolving Vectors into Axial Components and Vectors: Vector Addition.

Example Problem: A 2.0 kg ball, A\begin{align*}A\end{align*}, is moving with a velocity of 5.00 m/s due west. It collides with a stationary ball, B\begin{align*}B\end{align*}, also with a mass of 2.0 kg.  After the collision, ball A\begin{align*}A\end{align*} moves off at 30° south of west while ball B\begin{align*}B\end{align*} moves off at 60° north of west.  Find the velocities of both balls after the collision.

Solution: Since ball B\begin{align*}B\end{align*} is stationary before the collision, then the total momentum before the collision is equal to momentum of ball A\begin{align*}A\end{align*}.  The momentum of ball A\begin{align*}A\end{align*} before collision is shown in red below, and can be calculated to bep=mv=(2.00 kg)(5.00 m/s)=10.0 kg m/s west\begin{align*}p = mv = (2.00 \ \text{kg})(5.00 \ \text{m/s}) = 10.0 \ \text{kg m/s west}\end{align*}

Since momentum is conserved in this collision, the sum of the momenta of balls A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} after collsion must be 10.0 kg m/s west.

pAafter=(10.0 kg m/s)(cos30)=(10.0 kg m/s)(0.866)=8.66 kg m/s\begin{align*}p_{A \text{after}} = (10.0 \ \text{kg m/s})(\cos 30^\circ) = (10.0 \ \text{kg m/s})(0.866) = 8.66 \ \text{kg m/s}\end{align*}

pBafter=(10.0 kg m/s)(cos60)=(10.0 kg m/s)(0.500)=5.00 kg m/s\begin{align*}p_{B \text{after}} = (10.0 \ \text{kg m/s})(\cos 60^\circ) = (10.0 \ \text{kg m/s})(0.500) = 5.00 \ \text{kg m/s}\end{align*}

To find the final velocities of the two balls, we divide the momentum of each by its mass. Therefore, vA=4.3 m/s\begin{align*}v_A = 4.3 \ \text{m/s}\end{align*} and vB=2.5 m/s\begin{align*}v_B = 2.5 \ \text{m/s}\end{align*}.

Example Problem: A 1325 kg car moving north at 27.0 m/s collides with a 2165 kg car moving east at 17.0 m/s.  The two cars stick together after the collision.  What is the speed and direction of the two cars after the collision?

Solution:

Northward momentum=(1325 kg)(27.0 m/s)=35800 kg m/s\begin{align*}\text{Northward momentum} = (1325 \ \text{kg})(27.0 \ \text{m/s}) = 35800 \ \text{kg m/s}\end{align*}

Eastward momentum=(2165 kg)(17.0 m/s)=36800 kg m/s\begin{align*}\text{Eastward momentum} = (2165 \ \text{kg})(17.0 \ \text{m/s}) = 36800 \ \text{kg m/s}\end{align*}

R=(35800)2+(36800)2=51400 kgm/s\begin{align*}R=\sqrt{(35800)^2+(36800)^2}=51400 \ \text{kg} \cdot \text{m/s}\end{align*}

θ=sin13580051400=44 north of east\begin{align*}\theta=\sin^{-1} \frac{35800}{51400}=44^\circ \ \text{north of east}\end{align*}

velocity=pm=51400 kgm/s3490 kg=14.7 m/s @ 44 N of E\begin{align*}velocity=\frac{p}{m}=\frac{51400 \ \text{kg} \cdot \text{m/s}}{3490 \ \text{kg}}=14.7 \ \text{m/s} \ @ \ 44^\circ \ \text{N of E}\end{align*}

Example Problem: A 6.00 kg ball, A\begin{align*}A\end{align*}, moving at velocity 3.00 m/s due east collides with a 6.00 kg ball, B\begin{align*}B\end{align*}, at rest. After the collision, A\begin{align*}A\end{align*} moves off at 40.0° N of E and ball B\begin{align*}B\end{align*} moves off at 50.0° S of E.

1. What is the momentum of A\begin{align*}A\end{align*} after the collision?
2. What is the momentum of B\begin{align*}B\end{align*} after the collision?
3. What are the velocities of the two balls after the collision?

Solution: pinitial=mv=(6.00 kg)(3.00 m/s)=18.0 kg m/s\begin{align*}p_{\text{initial}} = mv = (6.00 \ \text{kg})(3.00 \ \text{m/s}) = 18.0 \ \text{kg m/s}\end{align*}

This is a right triangle in which the initial momentum is the length of the hypotenuse and the two momenta after the collision are the legs of the triangle.

1. pA=(18.0 kg m/s)(cos40.0)=(18.0 kg m/s)(0.766)=13.8 kg m/s\begin{align*}p_A = (18.0 \ \text{kg m/s})(\cos 40.0^\circ) = (18.0 \ \text{kg m/s})(0.766) = 13.8 \ \text{kg m/s}\end{align*}
2. pB=(18.0 kg m/s)(cos50.0)=(18.0 kg m/s)(0.643)=11.6 kg m/s\begin{align*}p_B = (18.0 \ \text{kg m/s})(\cos 50.0^\circ) = (18.0 \ \text{kg m/s})(0.643) = 11.6 \ \text{kg m/s}\end{align*}
3. vA=2.30 m/svB=1.93 m/s\begin{align*}v_A = 2.30 \ \text{m/s} \qquad v_B = 1.93 \ \text{m/s}\end{align*}

Summary

• The conservation of momentum law holds for all closed systems regardless of the directions of the objects before and after they collide.
• Momentum is a vector; collisions in two dimensions can be represented by axial vector components.

Practice

This video shows circus performers using conservation of momentum. Use this resource to answer the questions that follow.

1. Why do the fliers scrunch up in the air while spinning and twisting?
2. What happens to the rate at which they spin when they change shape in the air?

Review

1. Billiard ball A\begin{align*}A\end{align*}, mass 0.17 kg, moving due east with a velocity of 4.0 m/s, strikes stationary billiard ball B\begin{align*}B\end{align*}, also mass of 0.17 kg.  After the collision, ball A\begin{align*}A\end{align*} moves off at an angle of 30° north of east with a velocity of 3.5 m/s, and ball B\begin{align*}B\end{align*} moves off at an angle of 60° south of east. What is the speed of ball B\begin{align*}B\end{align*}?
2. A bomb, originally sitting at rest, explodes and during the explosion breaks into four pieces of exactly 0.25 kg each. One piece flies due south at 10 m/s while another pieces flies due north at 10 m/s.
1. What do we know about the directions of the other two pieces and how do we know it?
2. What do we know about the speeds of the other two pieces and how do we know it?
3. In a head-on collision between protons in a particle accelerator, three resultant particles were observed. All three of the resultant particles were moving to the right from the point of collision. The physicists conducting the experiment concluded there was at least one unseen particle moving to the left after the collision. Why did they conclude this?

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Date Created:
Feb 05, 2013